The proof that is given is short and based on an argument involving the determinant of a matrix. I don't find it though very insightful algebraically.
Does anybody know any alternative proof? I tried induction on the number of generators, but i have difficulty completing the induction step.

I should note that in Atiyah/McDonald, they use the word 'adjoint' in a way that we don't usually today. When they say 'adjoint,' they in fact mean the 'adjugate.' This makes their proof more meaningful (it confused me when I first read it)
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mixedmath♦Apr 17 '12 at 20:02

2 Answers
2

If $A$ is a field and $M$ has dimension $d$ its endomorphisms "are" $d\times d$ matrices with coefficients in $A$. This is a vector space of dimension $d^2$ over $A$ and since powers of $\phi$ are endomorphisms, only $d^2$ of them can be linearly independent, which gives you a linear relation between powers of $\phi$ such as that you wrote, taking e.g. $n=d^2+1$.

I guess that what Atiyah-Macdonald prove in their book is that the characteristic polynomial $\det(I-t\phi)$ is a particular such linear relation - the Cayley-Hamilton theorem. This polynomial is very useful to obtain information on $\phi$ and to relate linear algebra and polynomials in interesting ways. And this proof result holds for $A$ an arbitrary commutative ring. Commutative rings have the invariant basis number property but I think this does not imply that linearly independent sets in finitely generated modules cannot have infinite cardinality. So you cannot get a linear relation as for a field, where linearly independent sets have a maximal cardinality. So you are limited to the Cayley-Hamilton construction for an annihilator of $\phi$. But for most commutative rings you should have finiteness of linearly independent sets in f.g. modules, which implies linear relations between powers of endomorphisms.

Another remark: if $A$ is a field, $A[t]$ is a PID and since in general the linear relations between powers of $\phi$ form an ideal, there is a single polynomial, the minimal polynomial of $\phi$, generating all such linear relations.

Is the following the proof in Atiyah & Macdonald? I find it constructive:

The $A$-module $M$ is the quotient of $A^n$ for some natural number $n$.
There exists a morphism $\Phi: A^n \to A^n$ covering $\phi: M \to M$.
Since $A$ is a commutative ring, determinant makes sense and one can define the characteristic polynomial $p_\Phi(t) := \det(\Phi-t\,\mathrm{Id})$.
By the Cayley–Hamilton Theorem, $p_\Phi(\Phi)=0: A^n \to A^n$.
Observe that this morphism covers $p_\Phi(\phi): M \to M$.
Since the morphism $A^n \to M$ is surjective, it follows that $p_\Phi(\phi)=0$ also.

For a given $\phi$, I don't know anything about the uniqueness the polynomial $p_\Phi(t)$.